Question

An equation of the line tangent to the graph of f'(x) = (1-2x)³ + 3x(1-2x)² (-2). ƒ'(1) = (−1)³ + 3(−1)²(-2) = −7 OA. y = 7x-8 B. y=-2x+1 O C. y=2x-3 O D. y=-7x+6 O E. y=-6x + 5 at the point (1,-1) is:

Answer 1

hmmm let's reword all that mumble jumble.

we have a function graphed which is f(x), and we know its derivative which is f'(x), now, at the point (1 , -1) of f(x) f'(x) = -7, or we can say the instant slope when x = 1 is -7, so that m = -7 for f(x), what the heck is its equation?

well, we know its slope at x = 1, is -7, and we also have the point (1 , -1) for it, is all we need for its equation.

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\hspace{10em} \stackrel{slope}{m} ~=~ -7 \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{-7}(x-\stackrel{x_1}{1}) \implies y +1 = -7 ( x -1) \\\\\\ y +1 = -7 x +7 \implies {\Large \begin{array}{llll} y = -7 x +6 \end{array}}`